$\dfrac{ 4l + 2m }{ 6 } = \dfrac{ -8l + 4n }{ 3 }$ Solve for $l$.
Answer: Multiply both sides by the left denominator. $\dfrac{ 4l + 2m }{ {6} } = \dfrac{ -8l + 4n }{ 3 }$ ${6} \cdot \dfrac{ 4l + 2m }{ {6} } = {6} \cdot \dfrac{ -8l + 4n }{ 3 }$ $4l + 2m = {6} \cdot \dfrac { -8l + 4n }{ 3 }$ Reduce the right side. $4l + 2m = {6} \cdot \dfrac{ -8l + 4n }{ {3} }$ $4l + 2m = {2} \cdot \left( -8l + 4n \right)$ Distribute the right side $4l + 2m = {2} \cdot \left( -{8l} + {4n} \right)$ $4l + 2m = -{16}l + {8}n$ Combine $l$ terms on the left. ${4l} + 2m = -{16l} + 8n$ ${20l} + 2m = 8n$ Move the $m$ term to the right. $20l + {2m} = 8n$ $20l = 8n - {2m}$ Isolate $l$ by dividing both sides by its coefficient. ${20}l = 8n - 2m$ $l = \dfrac{ 8n - 2m }{ {20} }$ All of these terms are divisible by $2$ $l = \dfrac{ {4}n - {1}m }{ {10} }$